What do andi and ori do in this program?

Question

``` .global main # makes label "main" globally known .text # Instructions follow .align 2 # Align instructions to 4-byte words main: movi r16,0x47 # Load the hexadecimal value 41 # to register r16 loop: mov r4,r16 # Copy to r4 from r16 nop # (later changed to call hexasc) nop # (later changed to mov r4,r2) movia r8,putchar # copy subroutine address to a register callr r8 # call subroutine via register addi r16, r16,1 # Add 1 to register r16 andi r16, r16, 0x7f # mask with 7 bits ori r16, r16, 0x20 # set a bit to avoid control chars br loop # Branch to loop .end # The assembler will stop reading here foo bar bletch # comes after .end - ignored ``` I can understand everything, I think, except how the two instructions `andi` and `ori` work in this case. ori appears to make so that the ASCII 20 positions forward is printed but why and how?

Accepted Answer

andi and ori are both bitwise operators:

To see the difference, concider"

```
and $rd, $rs, $rt
or  $rd, $rs, $rt 
```

versus

```
andi $rt, $rs, immed
ori  $rt, $rs, immed
```

http://www.cs.umd.edu/class/sum2003/cmsc311/Notes/Mips/bitwise.html

What do andi and ori do in this program?

Solution

Problem